Back to Questions
EEJ MAIN Chemistry
QUESTION #7112
Question 1
In a galvanic cell represented by $Pt(s) | H_{2}(g, 1\text{ bar}) | HCl(aq) | AgCl(s) | Ag(s)$, the potential is $0.92\text{ V}$ at a specific concentration. If the standard reduction potential for the silver-silver chloride electrode is $0.22\text{ V}$, what is the $pH$ of the $HCl$ solution? (Assume $\frac{2.303RT}{F} = 0.06$)
Correct Answer Explanation
The cell reaction is $\frac{1}{2}H_{2} + AgCl \rightarrow Ag + H^{+} + Cl^{-}$.
$E_{cell} = E^{0}_{cell} - 0.06 \log([H^{+}][Cl^{-}])$. Since $E^{0}_{H_{2}/H^{+}} = 0$, $E^{0}_{cell} = 0.22\text{ V}$.
$0.92 = 0.22 - 0.06 \log([H^{+}]^2)$ (assuming $[H^{+}] = [Cl^{-}]$).
$0.70 = -0.12 \log[H^{+}] = 0.12 \times pH$.
$pH = 0.70 / 0.06 \approx 11.6$.
$E_{cell} = E^{0}_{cell} - 0.06 \log([H^{+}][Cl^{-}])$. Since $E^{0}_{H_{2}/H^{+}} = 0$, $E^{0}_{cell} = 0.22\text{ V}$.
$0.92 = 0.22 - 0.06 \log([H^{+}]^2)$ (assuming $[H^{+}] = [Cl^{-}]$).
$0.70 = -0.12 \log[H^{+}] = 0.12 \times pH$.
$pH = 0.70 / 0.06 \approx 11.6$.
Sign in to join the conversation and share your thoughts.
Log In to Comment